Let $M$ be a smooth manifold. One $k$-form is a section of the bundle $\bigwedge^k T^\ast M$, that is, if $p\in M$ and $\omega$ is a $k$-form then $\omega_p$ is one $k$-linear alternating real function defined in the cartesian product of $k$ copies of the tangent space at $p$.
Given $\omega$ one can then build a new differential $k+1$ form called the exterior derivative $d\omega$. This can be easily defined in a coordinate system $(x,U)$ as
$$d\omega = \sum_{}d\omega_{i_1\dots i_k}\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}$$
where $\omega_{i_1\dots i_k}$ are the components of $\omega$ in the coordinate system $(x,U)$. One then can easily show this definition doesn't depend on the coordinate system chosen. Based on that it is possible to show many properties of the exterior derivative then.
Now, the problem is that I simply can't get one geometrical intuition on what the exterior derivative is. As I know, a $k$-form can be thought of as an object which performs measurements on $k$-vectors, that is, an object capable of measuring projections of $k$-vectors. This point of view comes from the fact that $\bigwedge^k V^\ast$ is isomorphic to the space of linear functions on $\bigwedge^k V$.
In that setting, what is the geometrical intuition behind the exterior derivative? Given one $k$-form how can one understand from one intuitive point of view what its exterior derivative is?
You can visualize the exterior derivative in terms of a boundary operation. The fact that $d \circ d = 0$ can then be understood as following from the fact that the boundary of a boundary is empty. The idea is difficult to explain without pictures, so here's a short PDF that explains the idea and gives some nice diagrams to help you visualize differential forms and the exterior derivative operation in the the Euclidean setting.
There's also a good discussion on mathoverflow on this topic which you might find useful.